The value of k for which the following lines
x-y-1=0
2x+3y-12=0
2x-3y+k=0
are concurrent is

The lines are concurrent, so area of triangle formed by the lines will be zero.

So, $$\begin{vmatrix}1&-1 & -1 \\ 2 &3&-12\\ 2 & -3 & k \end{vmatrix}$$=0

Applying transformation, C2-->C1+C2 and C3-->C3+C1,

$$\begin{vmatrix}1&0 & 0 \\ 2 &5&-10\\ 2 & -1 & (k+2) \end{vmatrix}$$=0

Expanding along R1,

$$1\cdot$$ $$\begin{vmatrix}5 & -10 \\ -1&(k+2)\end{vmatrix}$$=0

So, $$5\left(k+2\right)-\left(-1\right)\cdot\left(-10\right)=0$$

or, $$5k+10-10=0$$

or, $$5k=0$$

or, $$k=0$$